Question

If $k$ is given a real constant $f\left(x\right)$ is a real quadratic in '$x$ such that $f(x+k)=f(-x)$. The coefficient of $x^2$ is $1$. Find $f\left(x\right)$.

Answer 1

Let $f\left(x\right)=x^2+bx+c$

$f(k+x)=f(-x){(x+k)}^2+b(x+k)+c={(-x)}^2+b(-x)+c{x}^2+k^2+2kx+bx+bk+c=x^2-bx+c{x}^2+(2k+b)x+(k^2+bk+c)=x^2-bx+c$

Comparing both sides
$\Rightarrow 2k=-2b\Rightarrow b=-k\Rightarrow f\left(x\right)=x^2-kx+c$

where $c\in R$